Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 1}{r^2 - 9r + 20} \div \dfrac{-4r + 4}{-4r + 20} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{r - 1}{r^2 - 9r + 20} \times \dfrac{-4r + 20}{-4r + 4} $ First factor the quadratic. $q = \dfrac{r - 1}{(r - 5)(r - 4)} \times \dfrac{-4r + 20}{-4r + 4} $ Then factor out any other terms. $q = \dfrac{r - 1}{(r - 5)(r - 4)} \times \dfrac{-4(r - 5)}{-4(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 1) \times -4(r - 5) } { (r - 5)(r - 4) \times -4(r - 1) } $ $q = \dfrac{ -4(r - 1)(r - 5)}{ -4(r - 5)(r - 4)(r - 1)} $ Notice that $(r - 1)$ and $(r - 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -4(r - 1)\cancel{(r - 5)}}{ -4\cancel{(r - 5)}(r - 4)(r - 1)} $ We are dividing by $r - 5$ , so $r - 5 \neq 0$ Therefore, $r \neq 5$ $q = \dfrac{ -4\cancel{(r - 1)}\cancel{(r - 5)}}{ -4\cancel{(r - 5)}(r - 4)\cancel{(r - 1)}} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $q = \dfrac{-4}{-4(r - 4)} $ $q = \dfrac{1}{r - 4} ; \space r \neq 5 ; \space r \neq 1 $